Yes! The above is totally correct that is because if we define $\displaystyle A(\varphi,\psi)=\left\{x:\varphi(x)=\psi(x)\right\ }$ and $\displaystyle \varphi\oplus\psi:X\to Y\times Y$ by $\displaystyle \varphi\oplus\psi:x\mapsto(\varphi(x),\psi(x))$ then the continuity of $\displaystyle \varphi,\psi$ easily implies the continuity of $\displaystyle \varphi\oplus\psi$. Thus, one notes that since $\displaystyle Y$ is Hausdorff that $\displaystyle \Delta\subseteq Y\times Y$ is closed and $\displaystyle A(\varphi,\psi)=\left(\varphi\oplus\psi\right)^{-1}(\Delta)$ and so $\displaystyle A(\varphi,\psi)$ is closed. So, if $\displaystyle D\subseteq \subseteq A(\varphi,\psi)\subseteq X$ is dense then $\displaystyle X=\overline{D}\subseteq\overline{A(\varphi,\psi)}= A(\varphi,\psi)$ and so $\displaystyle X=A(\varphi,\psi)$. But! By definition this means that $\displaystyle \varphi(x)=\psi(x),\text{ }\forall x\in X$ or that $\displaystyle \psi=\varphi$!

Maybe more intuitive is to notice that $\displaystyle \mathbb{R}$ is a topological group (with the usual addition and usual topology). And since given a topological group $\displaystyle G$ the map $\displaystyle \alpha:G\times G\to G$ given by $\displaystyle \alpha

g,h)\mapsto gh^{-1}$ is continuous. Thus, one notes that $\displaystyle \varphi\oplus\psi:G\to G\times G$ is continuous and thus so is $\displaystyle \alpha\circ(\varphi\oplus\psi):G\to G$ with $\displaystyle g\mapsto \varphi(g)(\psi(g))^{-1}$. So, noticing that $\displaystyle A(\varphi,\psi)=\left\{g\in G:\varphi(g)=\psi(g)\right\}=\left\{g\in G:\varphi(g)(\psi(g))^{-1}=e\right\}$$\displaystyle =\left(\alpha(\varphi\oplus\psi)\right)^{-1}(\{e\})$ and thus the above claim is true if $\displaystyle \{e\}$ is closed in $\displaystyle G$. And since $\displaystyle \{0\}$ is closed in $\displaystyle \mathbb{R}$ the conclusion follows!

(notice that the above paragraph is a farce since $\displaystyle \{e_G\}\text{ is closed }\implies G\text{ is }T_1\implies G\text{ is }T_2$)